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Studentka2010 [4]
2 years ago
7

Orchestra instruments are commonly tuned to match an A-note played by the principal oboe. The Baltimore Symphony Orchestra tunes

to an A-note at 440 Hz while the Boston Symphony Orchestra tunes to 442 Hz. If the speed of sound is constant at 343 m/s, find the magnitude of difference between the wavelengths of these two different A-notes. (Enter your answer in m.)
Physics
1 answer:
Nina [5.8K]2 years ago
6 0

Answer:

Δλ = 3*10⁻³ m.

Explanation:

  • At any wave, there exists a fixed relationship between the speed of  the wave, the wavelength, and the frequency, as follows:

       v = \lambda* f  (1)

       where v is the speed, λ is the wavelength and f is the frequency.

  • Rearranging terms, we can get λ from the other two parameters, as follows:

       \lambda = \frac{v}{f}  (2)

  • Since v is constant for sound at 343 m/s, we can find the different wavelengths at different frequencies, as follows:

        \lambda_{1} =\frac{v}{f_{1}} = \frac{343m/s}{440(1/s)} = 0.779 m  (3)

        \lambda_{2} =\frac{v}{f_{2}} = \frac{343m/s}{442(1/s)} = 0.776 m  (4)

  • The difference between both wavelengths, is just the difference between (3) and (4):

       \Delta \lambda = \lambda_{1} - \lambda_{2} = 0.779 m - 0.776m = 3e-3 m (5)

       ⇒ Δλ = 3*10⁻³ m.

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See the answer below

Explanation:

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As a wave moves through a medium, particles are displaced and (2 points)
Scorpion4ik [409]

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Explanation:

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7 0
3 years ago
A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an
Vlada [557]

Answer:

Explanation:

Given

volume V=16 ft^3

Suppose base is square with side L

height of crate is h

Volume V=L^2\times h

16=L^2\times h

Cost of top and bottom area c_1=3L^2

Cost of Side area c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}

Total Cost C=c_1+c_2

Total Cost C=3L^2+\frac{128}{L}

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Dimensions are L\times L\times h=2.75\times 2.75\times 11.46    

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